Solar Math

[jam]

Hello,

I've been scouring the net for information regarding the math behind a solar/wind system, unfortunately I couldn't find anything that would explain this simply.

For a beginner these seem to be a bunch of confusing concepts.

I'd like to take my home consumption as an example.
The electric bills says something like 1210KWh for the last 57 days. In turn this would make 21KWh per day, which in average would be 884W per hour.

In this case, would it be safe to assume that a 12V battery bank(10x 200Ah@20Hr 12V in //) providing 2000Ah rated @20Hr would provide about 1000W/h (2000Ah/24h * 12V ~= 1000W ) for 24h?

Another question: is there any way to measure the actual current consumption at any given time, other than relying on the utility bill reading (would a clamp AmpMeter on the electric line inlet do the trick)?

I'm guessing this would then go through a 12DC-120AC power inverter rated between 2KW and 5KW to power a house (given the consumption specs above).

Also, what is usually the efficiency of solar panels and how many would be needed to charge such a bank considering that at in Dec 21 (shortest day of the year) the daylight durations is 8h42m (at my current location).
What other factors should be taken into account?
What output should be expected (Voltage and charging current) of a single solar panel (say a 240W panel) under real conditions?

I'm not really interested in calculating costs, but more about the theoretical approach to solar power, off-the-grid systems and the considerations that need to be made for any given application (other than the actual cost).

Thank you!

[russ]

Hi jam - Welcome to Solar Panel Talk!

The battery math I and some other parts I will leave for others. Some of the points I can reply to.

1) Measuring individual items consumption. An amp clamp works fine for anything with a steady rate of consumption - say a light bulb or fan that runs 24 hours or that you know how many hours a day it operates.

2) A meter such as kill a watt (or equivalent) or a simple electronic power meter with plugs attached (what I have) that can be placed in line to an appliance will provide you with the amount used - a 24 hour period is far more accurate than less. The kill a watt type will give you a surge power reading as well I believe which my type of meter doesn't.

3) Items like a stove that is hardwired are a bit more difficult but you can work out with an amp clamp and estimated times.

4) Your location is important as far as insolation (amount of sunlight received in kW/m2/day. It has little to do with the 'daylight' hours reported (the 8 hours & 42 mins) . PV Watts provided by NREL will give you insolation figures for about any place around the world.

http://rredc.nrel.gov/solar/calculators/PVWATTS/version1/

For example, where I am we received 3.46 kW/m2 today - probably the best 4 hours were useful plus maybe 30 minutes on either side or about 3 kW/m2

If you provide your general location it would help.

5) Panel efficiency - There are a couple of different parts to any answer. Do you plan to be off grid or a hybrid system (batteries plus grid connect). It makes a big difference in the efficiencies you can obtain.

Russ

[jam]

Hi jam - Welcome to Solar Panel Talk!

The battery math I and some other parts I will leave for others. Some of the points I can reply to.

1) Measuring individual items consumption. An amp clamp works fine for anything with a steady rate of consumption - say a light bulb or fan that runs 24 hours or that you know how many hours a day it operates.

2) A meter such as kill a watt (or equivalent) or a simple electronic power meter with plugs attached (what I have) that can be placed in line to an appliance will provide you with the amount used - a 24 hour period is far more accurate than less. The kill a watt type will give you a surge power reading as well I believe which my type of meter doesn't.

3) Items like a stove that is hardwired are a bit more difficult but you can work out with an amp clamp and estimated times.

4) Your location is important as far as insolation (amount of sunlight received in kW/m2/day. It has little to do with the 'daylight' hours reported (the 8 hours & 42 mins) . PV Watts provided by NREL will give you insolation figures for about any place around the world.

http://rredc.nrel.gov/solar/calculators/PVWATTS/version1/

For example, where I am we received 3.46 kW/m2 today - probably the best 4 hours were useful plus maybe 30 minutes on either side or about 3 kW/m2

If you provide your general location it would help.

5) Panel efficiency - There are a couple of different parts to any answer. Do you plan to be off grid or a hybrid system (batteries plus grid connect). It makes a big difference in the efficiencies you can obtain.

Russ

Thank you for the welcome and for the reply.

I am looking at the figures for an off-grid implementation, but also trying to estimate a hybrid system (grid+solar).
I'd like to know what to expect from both.

According to the site you provided, in the worst case scenario I get 2.27kWh/m2/day at my current location.

My focus right now is only on the theory and I'd like to get my numbers right.

Thank you!

[russ]

You are doing the right thing - study well before leaping into expense.

All types of systems are expensive enough that you will likely be stuck with your first purchase - so you need to get it right!

Let's see other comments about batteries etc.

Russ

[Sunking]

According to the site you provided, in the worst case scenario I get 2.27kWh/m2/day at my current location.

Ok that translates to 2.27 Sun Hours which is the number you would have to use for a Stand Alone Battery system.

So using the 21 Kwh per day you quoted here is how things work out.

For this large of a system you are looking at minimum a 48 volt battery system. So at 48 volts the battery size is (21,000 wh x 5) / 48 volts = 2188 Amp Hours. Round that up to 2200 Amp Hours at 48 volts. That works out to about 6300 pounds of battery.

For the solar panel wattage the best efficiency you can expect using a MPPT charge controller is 66%. 20 to generate 21 Kwh of usable power your panels need to generate 21,000 / .66 = 31,800 Kwh per day. So with a winter 2.27 Sun Hours you would need 31800 wh / 2.27 hours = 14,000 watt solar panel array.

If you want to play with battery numbers go here and see what it takes.

That should get you started.

[jam]

Ok that translates to 2.27 Sun Hours which is the number you would have to use for a Stand Alone Battery system.

So using the 21 Kwh per day you quoted here is how things work out.

For this large of a system you are looking at minimum a 48 volt battery system. So at 48 volts the battery size is (21,000 wh x 5) / 48 volts = 2188 Amp Hours. Round that up to 2200 Amp Hours at 48 volts. That works out to about 6300 pounds of battery.

For the solar panel wattage the best efficiency you can expect using a MPPT charge controller is 66%. 20 to generate 21 Kwh of usable power your panels need to generate 21,000 / .66 = 31,800 Kwh per day. So with a winter 2.27 Sun Hours you would need 31800 wh / 2.27 hours = 14,000 watt solar panel array.

If you want to play with battery numbers go here and see what it takes.

That should get you started.

Thanks for the reply, that does clarify some things.

I am not sure however why you state I should go with 48V as opposed to 12V (I'm guessing the total amount of batteries would be the same, with only the //-series setups changing.)

What difference would it make to actually have a higher voltage with a lower current rather than having a lower voltage with a high current?

Thank you!

[Sunking]

What difference would it make to actually have a higher voltage with a lower current rather than having a lower voltage with a high current?

Money and lots of it, and efficiency.

High current = very large expensive copper conductors..

At the kind of power levels you are talking about I would be running 190 volts up to 400 volts on the batteries.